Chief Delphi - Maths problems (interesting ones)

Chief Delphi (http://www.chiefdelphi.com/forums/index.php)

- Math and Science (http://www.chiefdelphi.com/forums/forumdisplay.php?f=70)

- - Maths problems (interesting ones) (http://www.chiefdelphi.com/forums/showthread.php?t=29393)

Re: Maths problems (interesting ones)

Hi

for the coin question:

label coins a-l
measure 4 (a-d) coins against another 4(e-h)
if the same then the spare 4 (i-l) contains the odd one out, measure 1(i)v1(j) of these
if i and j are the same take one of the spares (k) and measure against one of those you know to be true (i or j), if the same, the unmeasured one (l) is the fake, if different kmust be fake
if the i and j are different you know any of the remaining coins are real so measure i against any of the others to know if it is fake, if it isnt j is.

if a-d are different to e-h (remember which is heavier)
take two from each set a,b,e,f and measure the against i-l if they are different you can work out whether the fake is heavier or lighter; which means you know whether a or b are fake or e or f are fake, you can then eliminate which one in the same method as i or j above

if they are the same take c and g and measure against i and j
if they are different, you know whether the fake is heavier or lighter so can work out whether it is c or g that is the fake
if they are the same mesure one of them against a real coin and find which one is fake in the same way as i and j above

now for another coin question:
there are a thousand coins lined up in a row, all with heads up. If somone went along and turned over every coin, and when they finished that someone else came along and turned over every second, and when that was done someone turned over every third, and so on up to a thousand how many coins would be facing heads up when it was finished

oljomo

Re: Maths problems (interesting ones)

Quote:

Originally Posted by oljomo

Hi

for the coin question:

label coins a-l
measure 4 (a-d) coins against another 4(e-h)
if the same then the spare 4 (i-l) contains the odd one out, measure 1(i)v1(j) of these
if i and j are the same take one of the spares (k) and measure against one of those you know to be true (i or j), if the same, the unmeasured one (l) is the fake, if different kmust be fake
if the i and j are different you know any of the remaining coins are real so measure i against any of the others to know if it is fake, if it isnt j is.

if a-d are different to e-h (remember which is heavier)
take two from each set a,b,e,f and measure the against i-l if they are different you can work out whether the fake is heavier or lighter; which means you know whether a or b are fake or e or f are fake, you can then eliminate which one in the same method as i or j above

if they are the same take c and g and measure against i and j
if they are different, you know whether the fake is heavier or lighter so can work out whether it is c or g that is the fake
if they are the same mesure one of them against a real coin and find which one is fake in the same way as i and j above

now for another coin question:
there are a thousand coins lined up in a row, ll with heads up. If somone went along and turned over every coin, and when they finished that someone else came along and turned over every second, and when that was done someone turned over every third, and so on up to a thousand how many coins would be facing heads up when it was finished

oljomo

Hey you brits are smart! congratulations.
I probably won't have to post the 2 solutions with 4 ways I found?!
So now it's your turn!

Re: Maths problems (interesting ones)

Hi

Another problem:

ian and jenny both have a whole amount of pounds (im british)
Ian says to Jenny "if you give me £3 i will have n times as much as you"
Jenny says to Ian "if you give me £n i will have 3 times as much as you"
Given that n is an integer find the possible values for n

oljomo

Re: Maths problems (interesting ones)

Quote:

Originally Posted by oljomo

OK.. Here goes...

If ian has £x and Jenny has £y, then 3+x=n(y-3) and n+y=3(x-n)
use equation 2 and y=3x-3n
use equation 1 and 3+x=3xn-4n^2-3n
solving equation 1 for 0, 0=3xn-4n^2-3n-x-3
(you could work the other way with equation 2 and get to this same point)
pluging in some numbers for x (3,4,5,6,7,8,9,10) you get n=1 or 2 for x=5 and n=3 or .75 for x=6 and the rest are very odd or imaganary.

Therefore n=1,2 or 3

Re: Maths problems (interesting ones)

Quote:

Originally Posted by unapiedra

ok, here got it from a book some place and i figured out the solution (actually there are two but only two)

--so here it goes: a 9-digit number (123456789; the digits only represent the position for further labeling),
--now each position has to be divideable (english, gramar, I sorry.) by the #of the position (like 1/1=n; 12/2=n; 123/3=n; 1234/4=n and so on; n representing a natural number)
--and finaly, the number consist of the digits 1 through 9, but they can be used only once

So get the solution, and I'm sorry for my terrible english

Hi

solution is

381654729

jdiwnab, you have not got all the answers.

oljomo

Re: Maths problems (interesting ones)

Contenuing on with the numbers from the above question: and then 7 (if there are more, it is not in the range of 3-147)

Re: Maths problems (interesting ones)

Lets wrap up the unsolved puzzles.

11)

Quote:

Originally Posted by mtrawls

Suppose that there is a game where we flip a coin. If the first time the coin is flipped it lands on heads, I pay you $2 (2^1), if the first time it lands on heads is the second toss, I pay you $4 (2^2), if the first time it lands on heads is the third toss, I pay you $8 (2^3), ad infinitum. Using simple probability, tell me how much you would pay me to play this game and still be able to make a decent profit (but enough to tempt me to agree to the game).

Here are your chances of winning amounts of money:

50%.......0$
25%.......2$
12.5%....4$
6.25%....8$
3.125%..16$
etcetera

The average winnings of the game, is therefore: 0*(0.5)+2*(0.25)+4*(0.125)+8*(0.0625)+16*(0.03125) ...
This infinite series yields, amazingly, infinity. Therefore, if my math is correct, I could pick an arbitrarily large number of games, pay you whatever you wanted, (lets say 3 million games for 10 million dollars), and I would come out ahead.
Yet, am not certain in this result, so I would stick to paying a dollar per game.

Short answer: Any amount?

Another thing: if you were to stick to your payment scheme, you would pay me a dollar even if I loose the first round. However, this does not effect the results.

14)

Quote:

Originally Posted by Denman

Let us go into heroic fantasy land. The story takes place in a antic land where dwarves are mining to extract gold. The king is visiting the mine when he ears voices around the corner: "I have a neat trick: we have to melt gold bars the size of this mold, that is exactly 100 grams. I remove just a tiny fraction to all my bars and they all weight 99 grams. Their scales are not precise enough, they don't see anything..." The king tries to run after then, but they ear him and manage to flee among the corridors. Annoyed, the king summons the gods, who grant him a gift from the future: a digital scale. The digital scales works as follows: you put whatever you want on the scale, (nothing happens then), you press a button and the exact weight of what is on the scale appears. Then, the scale vanishes. Note that before you have pressed the button, no weight appears and once you have pressed the button the weight won't change even if you add or remove things. Armed with this mighty scale, the king summons all his 100 workers. How will he find the cheater?

This problem requires that you can take portions of each person's supply. So take 1 bar of the first person's supplies, 2 bars of the next person, 3 bars of the next, and so on until you have some of every worker's suplies on the scale. When the scale is read, you can then determine which person cheated based on how much off the scale was to the ideal amount. In this example, with 100 workers and starting with 100 bars, 505000 grams - Measured grams = number of person who cheated.

17)

Quote:

Originally Posted by Denman

Divide $261 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $261, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

Have bags that are filled with 1$, 2$, 4$, 8$, 16$, 32$, 64$, 128$, $6
That is 9 bags.

This is what the binary system is all about; the minimum number of on/off digits to get any value. With 8 bags, you can get any number from 0 to 255. With 9, you can go from 0 to 511.

19)

Quote:

Originally Posted by oljomo

there are a thousand coins lined up in a row, all with heads up. If someone went along and turned over every coin, and when they finished that someone else came along and turned over every second, and when that was done someone turned over every third, and so on up to a thousand how many coins would be facing heads up when it was finished

31 will be be tails
969 will be heads

All those numbers that have an even number of factors will be heads at the end. But those that have an odd number of factors will end up tails. The only way a number can have an odd number of factors is for a factor to repeat itself, like 16 has 1, 2, 4, 8, and 16. Because the factor repeats itself, the number is a square of an interger. There are 31 squares of intergers between 1 and 1000 (31 squared being 961 and 32 squared being 1024).

And now for some new ones!

21)
1000! (1000 factorial) has how many zeros at the end?

22)
Which is bigger and by how much? 1000 to the 1001 power or 1001 to the 1000 power? Once you get the solution, can you prove that one divided by the other is close to a famous irrational number?

Sorry for the long post.

Re: Maths problems (interesting ones)

Quote:

Originally Posted by kirish

Lets wrap up the unsolved puzzles.

17)

Have bags that are filled with 1$, 2$, 4$, 8$, 16$, 32$, 64$, 128$, $256
That is 9 bags.

This is what the binary system is all about; the minimum number of on/off digits to get any value. With 8 bags, you can get any number from 0 to 255. With 9,

you can go from 0 to 511.

I think you made a mistake right there.-- You only have $261, so you have to divide it some how different (even though I think that is impossible).
Or maybe I understood the problem wrong.

Quote:

And now for some new ones!

21) 1000! (1000 factorial) has how many zeros at the end?

I think that would be 700 zeros at the end. (1000/2) + (1000/5) = 700

Great job on the other ones!

Re: Maths problems (interesting ones)

Quote:

Originally Posted by oljomo

Hi

solution is

381654729

jdiwnab, you have not got all the answers.

oljomo

Neither have you! There is a second solution to it... Very similiar!

Re: Maths problems (interesting ones)

Quote:

Originally Posted by kirish

21)
1000! (1000 factorial) has how many zeros at the end?

22)
Which is bigger and by how much? 1000 to the 1001 power or 1001 to the 1000 power? Once you get the solution, can you prove that one divided by the other is a famous irrational number?

Hi

for the first one, it should have a 0 for every multiple of 10, 100 and also one for every multiple of five that can be coupled with an even number, (with so many thats going to be all of them), another hundred, if you don't count the multiples of ten. This should mean that there are 200 zeros at the end (i think)

For the second answer surely the question is contradictory, if its an irrational number by definition it cant be represented as a fraction (or at least i thought)
so the division cannot represent an irrational number.

1000^1001=10^3003
(1000+1)^1000=10^3000+1000*10^2997+500500*10^2994+ ..........+500500*10^6+1000*10^3+1 (by binomial distributions (only doing the first and last three you can see that the numbers get smaller as it goes on
=10^3000+10^3000+5*10^2994+........+5*10^11+10^6+1
which must be less than 10^3003

Therefore 1000^1001 is bigger

And for the money splitting up one unapiedrs point is easily overcome if you take the 256 bag and replace it with a 1, 2 and 2 bag, so you have 2 $1 bags and 3 $" bags in total) then you use $261, and can give out any amount.

And i cannot find the other solution (im sure ive tried all the ones that might work ad they fall down at the seven), so can you tell me the other answer

oljomo

Re: Maths problems (interesting ones)

Quote:

Originally Posted by unapiedra

I think you made a mistake right there.-- You only have $261, so you have to divide it some how different (even though I think that is impossible).
Or maybe I understood the problem wrong.

I think that would be 700 zeros at the end. (1000/2) + (1000/5) = 700

Great job on the other ones!

your right unapiedra, you only have 261 to start with thus you can't do that method.
Its nearly right though

Quote:

Originally Posted by oljomo

And for the money splitting up one unapiedrs point is easily overcome if you take the 256 bag and replace it with a 1, 2 and 2 bag, so you have 2 $1 bags and 3 $" bags in total) then you use $261, and can give out any amount.

oljomo

nearly
but you can get it less than that.

Re: Maths problems (interesting ones)

Quote:

Originally Posted by Denman

your right unapiedra, you only have 261 to start with thus you can't do that method.
Its nearly right though

nearly
but you can get it less than that.

hi

silly me you can get away with just a single $5 bag

oljomo

Re: Maths problems (interesting ones)

Sorry about the irrational number statement, I meant "close to" an irrational number. It is fixed it now.

Also, I fixed my answer to the money bag problem, so it only takes $261.

I am not sure yet what the answer for the factorial is, but I will have it soon. As for the exponent problem, I think you may need to use logs to get the "how many times bigger" part.

Also, I wrote a javascript program to find out the average winnings of the coin game (11). The answer seems to be between $5 and $15, but even with 5000 games played, it varies a lot. To run the program yourself, save the txt as a .html file and open it in your web browser.

Re: Maths problems (interesting ones)

Hi

for the log question

1000^1001/1001^1000
=e^ln(1000^1001)/e^ln(1001^1000)
=e^ln(1000^1001)-ln(1001^1000)
=e^1000ln(1000)-1000ln(1001)+ln(1000)
=e^1000ln(1000/1001)+ln(1000)
=368.0633043

therefore one is 368 times bigger than the other
(the reciprocal
is 0.0027169, which is near to e\1000

oljomo